Project Euler - Problem 28

最新推荐文章于 2025-12-26 14:14:13 发布

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#include #ios

本文提供两段C++代码示例，分别用于计算特定对角线数值之和并展示运行效率。第一段代码计算两个方向的对角线数值总和，第二段代码则通过迭代增加的方式计算直至1000的所有对角线数值总和，并使用clock函数测量程序运行时间。

#include <iostream> #include <time.h> #include <list> using namespace std; int main() { double start = clock(),end(0); unsigned int add1 = 1; unsigned int sum1 = 0; //计算/方向对角线和 for(unsigned int a=2; a<11; a+=2) { add1 += a; sum1 += add1; //cout << add <<endl; } //计算/方向对角线和 unsigned int add2 = 1; unsigned int sum2 = 0; for(unsigned int b=4; b<11; b+=4) { int count = 1; while( count<3 ) { add2 += b; sum2 += add2; ++count; } } cout << endl << sum1+sum2-1; end = clock(); double result = (end - start)/1000; cout.setf(ios::fixed); cout <<endl<< "3.00GHz's run time = " << result << "s."<<endl; return 0; }

#include <iostream> #include <time.h> #include <list> using namespace std; int main() { double start = clock(),end(0); unsigned int diff = 1; unsigned int sum = 1; for(unsigned int t=2; t<1001; t+=2) { int count = 0; while(count < 4) { diff += t; sum += diff; count++; } } cout << "the sum is " <<sum; end = clock(); double result = (end - start)/1000; cout.setf(ios::fixed); cout <<endl<< "3.00GHz's run time = " << result << "s."<<endl; return 0; }